I want to build a right triangle, 30 deg. 60 deg. angles without using a
protractor (don't have one). The long arm will be 24 inches and will be
using a square for the right angle. Anyone know the formula to get the
length of the short side? I can figure out the hypotenuse for a double
check of squareness.

Geometry is in my misty past by over 50 years.
--
Gerald Ross
Cochran, GA

"I am a marvelous housekeeper. Every
time I leave a man I keep his house."
Zsa Zsa Gabor





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Gerald Ross wrote:
I want to build a right triangle, 30 deg. 60 deg. angles without using a
protractor (don't have one). The long arm will be 24 inches and will be
using a square for the right angle. Anyone know the formula to get the
length of the short side? I can figure out the hypotenuse for a double
check of squareness.

Geometry is in my misty past by over 50 years.

The short side is long side/2

The medium side is short side x square root of 3.

These only apply for 30-60-90 triangles.

For your situation, the shortest side is 24"/1.732 = 13.857

On Sun, 09 Apr 2006 09:30:08 -0400, Gerald Ross wrote:

I want to build a right triangle, 30 deg. 60 deg. angles without using a
protractor (don't have one). The long arm will be 24 inches and will be
using a square for the right angle. Anyone know the formula to get the
length of the short side? I can figure out the hypotenuse for a double
check of squareness.

Geometry is in my misty past by over 50 years.

Short side is 1x, hypotenuse is 2x, and long side is 1.732x (sqroot of 3).
So your short side is a little less than 13 7/8.

D.G. Adams

On Sun, 09 Apr 2006 09:30:08 -0400, Gerald Ross
wrote:

I want to build a right triangle, 30 deg. 60 deg. angles without using a
protractor (don't have one). The long arm will be 24 inches and will be
using a square for the right angle. Anyone know the formula to get the
length of the short side? I can figure out the hypotenuse for a double
check of squareness.

13.856"


Geometry is in my misty past by over 50 years.

That's okay, it's trigonometry you needed anyway :-)

Taking the 30 degree angle, find the tangent and multiply it by 24"

Without trig tables or a calculator that can do trig, tan(30) = 1/3 *
sqrt(3) and everybody knows the sqrt(3) = 1.73, right?

Of if you want to consider the 60 degree angle first, the tan =
sqrt(3), then the short side = 24 / 1.73.

On Sun, 09 Apr 2006 08:35:04 -0500, Dave Nay
davedotnay@vidanaydotcom wrote:

Gerald Ross wrote:
I want to build a right triangle, 30 deg. 60 deg. angles without using a
protractor (don't have one). The long arm will be 24 inches and will be
using a square for the right angle. Anyone know the formula to get the
length of the short side? I can figure out the hypotenuse for a double
check of squareness.

Geometry is in my misty past by over 50 years.

The short side is long side/2

The medium side is short side x square root of 3.

These only apply for 30-60-90 triangles.

For your situation, the shortest side is 24"/1.732 = 13.857

That's possibly the way I'd do it, thanks to Pythagoras ...or go out
and buy a protractor, or compasses. I prefer compasses [or a trammel]
for specific angles like that [you can have, or make large ones and
get better accuracy over distance].

Or, having a computer to be able to post here, there is CAD. Print a
sheet with the angle, fold along the lines and use as a template.

Or there is a way to trisect an angle [approximate, but not that you'd
notice] using the square. So a 30 degree angle is possible. All
physical measures are inherently approximate even if not in theory, so
it would do.

Gerald Ross wrote:
I want to build a right triangle, 30 deg. 60 deg. angles without using a
protractor (don't have one). The long arm will be 24 inches and will be
using a square for the right angle. Anyone know the formula to get the
length of the short side? I can figure out the hypotenuse for a double
check of squareness.

Geometry is in my misty past by over 50 years.

Thanks for all the help. This is to make a rough triangle to check on
building a one-time project. Not a high accuracy carpentry job this time.

--
Gerald Ross
Cochran, GA

If we do not succeed, we run the risk
of failure. -- Dan Quayle





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On Sun, 09 Apr 2006 06:59:45 -0700, Wes Stewart
wrote:

Geometry is in my misty past by over 50 years.

That's okay, it's trigonometry you needed anyway :-)

Why not use geometry? For a 30° angle you can lay it out more easily
by geometry.

"you can lay it out more easily by geometry."

Not sure what you mean, Andy.

Perhaps you would care to post your solution here?

My searching seemed to indicate it was a Trig problem. But, then, I could
NOT find a solution worth posting. If you did, share.


"Andy Dingley" wrote in message
...
On Sun, 09 Apr 2006 06:59:45 -0700, Wes Stewart
wrote:

Geometry is in my misty past by over 50 years.

That's okay, it's trigonometry you needed anyway :-)

Why not use geometry? For a 30° angle you can lay it out more easily
by geometry.

On Mon, 10 Apr 2006 03:16:56 +0000, Gooey TARBALLS opined:

Perhaps you would care to post your solution here?

Draw a long line (1). Construct a perpendicular (2). Set dividers to
desired length of shortest leg (x). Strike that distance on the
perpendicular, starting at the intersection (origin, o). Step off twice
the divider setting on the first, horizontal line. Construct a
perpendicular at the 2x point (3). Strike the divider distance up the
second perpendicular (4). Connect the two upper points (5). (You now have
a 1x2 rectangle.) Set the dividers to 2x. Strike an arc which intersects
the upper horizontal (parens). Construct a third perpendicular to the
horizontal line which intersects the intersection of the second horizontal
and the 2x arc (6). Connect the origin to this last intersection (o, +).
QED.
2 4
x .------------+--.-- 5
| ) |
| |) |
| | )|
_o_______.____|__|__ 1
x 2x
6 3

--
"Keep your ass behind you"
wreck20051219 at spambob.net

"Australopithecus scobis" wrote in message
news
On Mon, 10 Apr 2006 03:16:56 +0000, Gooey TARBALLS opined:

Perhaps you would care to post your solution here?

Draw a long line (1). Construct a perpendicular (2). Set dividers to
desired length of shortest leg (x). Strike that distance on the
perpendicular, starting at the intersection (origin, o). Step off twice
the divider setting on the first, horizontal line.

SNIP etc...

Depending on age, he may not have studied any Euclidian geometry at all. No
proofs by construction in the last two "geometry" texts at our school.

Draw a vertical line desired length of shortest leg (x).
Draw a horizontal line 2x in length from the same point.

Could the result be that 30,60, 90 degree triangle? (if one connected the
points?


|
|
|
|______________________ (line one = 2X perpendicular)

This would seem to suggest that the formula for a 30, 60, 90 degree triangle
would be that side B = 2A and the hypotenuse = SQRT ((B*B)+(A*A))

Could it be this simple?



"Australopithecus scobis" wrote in message
news
On Mon, 10 Apr 2006 03:16:56 +0000, Gooey TARBALLS opined:

Perhaps you would care to post your solution here?

Draw a long line (1). Construct a perpendicular (2). Set dividers to
desired length of shortest leg (x). Strike that distance on the
perpendicular, starting at the intersection (origin, o). Step off twice
the divider setting on the first, horizontal line. Construct a
perpendicular at the 2x point (3). Strike the divider distance up the
second perpendicular (4). Connect the two upper points (5). (You now have
a 1x2 rectangle.) Set the dividers to 2x. Strike an arc which intersects
the upper horizontal (parens). Construct a third perpendicular to the
horizontal line which intersects the intersection of the second horizontal
and the 2x arc (6). Connect the origin to this last intersection (o, +).
QED.
2 4
x .------------+--.-- 5
| ) |
| |) |
| | )|
_o_______.____|__|__ 1
x 2x
6 3

--
"Keep your ass behind you"
wreck20051219 at spambob.net

Check this out:

http://home.comcast.net/~charliebcz/...Ratio/GM1.html


"Gerald Ross" wrote in message
...
Gerald Ross wrote:
I want to build a right triangle, 30 deg. 60 deg. angles without using a
protractor (don't have one). The long arm will be 24 inches and will be
using a square for the right angle. Anyone know the formula to get the
length of the short side? I can figure out the hypotenuse for a double
check of squareness.

Geometry is in my misty past by over 50 years.

Thanks for all the help. This is to make a rough triangle to check on
building a one-time project. Not a high accuracy carpentry job this time.

--
Gerald Ross
Cochran, GA

If we do not succeed, we run the risk
of failure. -- Dan Quayle





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News==----
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Newsgroups
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On Mon, 10 Apr 2006 15:07:38 GMT, "Gooey TARBALLS"
wrote:

Draw a vertical line desired length of shortest leg (x).
Draw a horizontal line 2x in length from the same point.

Could the result be that 30,60, 90 degree triangle? (if one connected the
points?

NO. As was pointed out the two arms [at right angles] of a 30,60
right triangle are in the ratio 1:sqrt(3). The Hypotenuse is the side
twice the shorter.

On Mon, 10 Apr 2006 05:30:35 GMT, Australopithecus scobis
wrote:

On Mon, 10 Apr 2006 03:16:56 +0000, Gooey TARBALLS opined:

Perhaps you would care to post your solution here?

Draw a long line (1). Construct a perpendicular (2).

etc..

If using straightedge and ruler, draw a horizontal line, and mark off
two equal distances AB, BC. From A draw an arc [no need to change
distances on the compasses] to cover half way [but above] from A to B.
using B, draw the same arc back towards A. Suppose the arc meet at D.
Draw a line through AD. Widen the compasses. Darw an arc from A
above B. Draw the same arc from C above B to meet the first in E.
Draw the line through BE.

Angle BAD will be 60 degrees. Angle ABE will be 90. There's your
30.60,90 triangle.

"Guess who" wrote in message ...
| On Mon, 10 Apr 2006 15:07:38 GMT, "Gooey TARBALLS"
| wrote:
|
| Draw a vertical line desired length of shortest leg (x).
| Draw a horizontal line 2x in length from the same point.
|
| Could the result be that 30,60, 90 degree triangle? (if one connected the
| points?
|
| NO. As was pointed out the two arms [at right angles] of a 30,60
| right triangle are in the ratio 1:sqrt(3). The Hypotenuse is the side
| twice the shorter.
|

You sure about this??

Seems to me the ratio of the sides of a scalene right triangle (30,60,90) is

1:2:sqrt3.

and, according to Pythagorus,

The square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other 2 sides.

Give all this balderdash, were one to lay out a horizontal base line of 2" and erect a vertical line at one end of the base line that is 1" in height, the distance between the two unattached line ends would be exactly sqrt 3" long. Additionally, the smallest angle would be exactly 30 degrees, the larger angle would be exactly 60 degrees and the biggest angle would be exactly 90 degrees.

--
PDQ

--

PDQ wrote:
"Guess who" wrote in message ...
| On Mon, 10 Apr 2006 15:07:38 GMT, "Gooey TARBALLS"
| wrote:
|
| Draw a vertical line desired length of shortest leg (x).
| Draw a horizontal line 2x in length from the same point.
|
| Could the result be that 30,60, 90 degree triangle? (if one connected the
| points?
|
| NO. As was pointed out the two arms [at right angles] of a 30,60
| right triangle are in the ratio 1:sqrt(3). The Hypotenuse is the side
| twice the shorter.
|

You sure about this??

Seems to me the ratio of the sides of a scalene right triangle (30,60,90) is

1:2:sqrt3.

and, according to Pythagorus,

The square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other 2 sides.

Give all this balderdash, were one to lay out a horizontal base line of 2" and erect a vertical line at one end of the base line that is 1" in height, the distance between the two unattached line ends would be exactly sqrt 3" long. Additionally, the smallest angle would be exactly 30 degrees, the larger angle would be exactly 60 degrees and the biggest angle would be exactly 90 degrees.

So, you're saying 1^2 + 2^2 = 3 ?

er
--
email not valid

Gooey TARBALLS wrote:
Draw a vertical line desired length of shortest leg (x).
Draw a horizontal line 2x in length from the same point.

Could the result be that 30,60, 90 degree triangle? (if one connected the
points?


|
|
|
|______________________ (line one = 2X perpendicular)

This would seem to suggest that the formula for a 30, 60, 90 degree triangle
would be that side B = 2A and the hypotenuse = SQRT ((B*B)+(A*A))

Could it be this simple?


The lengths of each arm are 1 and sqrt(3), the hypotenuse is length 2.
That was also given in austrobis' geometric proof.

er
--
email not valid

On Mon, 10 Apr 2006 14:13:23 -0400, "PDQ" wrote:

"Guess who" wrote in message ...
| On Mon, 10 Apr 2006 15:07:38 GMT, "Gooey TARBALLS"
| wrote:
|
| Draw a vertical line desired length of shortest leg (x).
| Draw a horizontal line 2x in length from the same point.
|
| Could the result be that 30,60, 90 degree triangle? (if one connected the
| points?
|
| NO. As was pointed out the two arms [at right angles] of a 30,60
| right triangle are in the ratio 1:sqrt(3). The Hypotenuse is the side
| twice the shorter.
|

You sure about this??

The square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other 2 sides.

and....???

On Mon, 10 Apr 2006 03:16:56 GMT, "Gooey TARBALLS"
wrote:

Perhaps you would care to post your solution here?

In ASCII ? Not really !

My searching seemed to indicate it was a Trig problem.

You can do it either way. For a general solution to any angle, then trig
would be the way to go - especially with cheap calculators. However
90°. 60° and 30° are easy constructions with just dividers and they're
worth knowing.

Draw a circle. With your dividers set on the same radius, step out this
distance repeatedly around the circumference of the circle. It fits
exactly 6 times, meaning that the radii to each of these points make
angles of 60°.

To get from 60° to 30°, then you can either bisect it or you can
subtract 60° from 90° (you'll often already have a right angle laid out
for some other purpose).

To bisect the angle, use the dividers as before and mark out a point at
approximately two radii from the centre, at the intersection of arcs
drawn from two of the circumferential points you marked earlier. A
radius drawn out to this point bisects the circumferential points, at an
angle of 30°

In article ,
Australopithecus scobis wrote:
On Mon, 10 Apr 2006 03:16:56 +0000, Gooey TARBALLS opined:

Perhaps you would care to post your solution here?

Draw a long line (1). Construct a perpendicular (2). Set dividers to
[[.. munch rest of a detailed description ..]]

*SIGH* There's an even easier way.

1) draw a straight line
2) label two points on it, 'A' and 'B', and set a compass to that measure.
3) draw a 180-degree arc, centered on A above the line.
4) draw another arc, centered on B, that intersects the arc in 3.
call that point of intersection C.
5) draw 'yet another' arc, centered on C, that intersects the arc in 3.
call that point of intersection D.
6) set the compass to something 'significantly more' than the AB measure
(how much more _doesn't_matter_, although the bigger you make it,
the less opportunity there is for 'human error' to creep in.)
7) using that larger setting, strike an arc from each end of the arc in 3
they will intersect somehere above the line that A and B are on
call that point of interesection E
8) draw a line AE, and extend it, if needed until it intersects the BD
line. call that point F.

ABF is a 30-60-90 right triangle.

It takes _way_ longer to describe it than to do it. grin

note: steps 6-7-8 could be reduced to 'strike a perpendicular'.

Nope.

"PDQ" wrote in message
...
Give all this balderdash, were one to lay out a horizontal base line of 2"
and erect a vertical line at one end of the base line that is 1" in height,
the distance between the two unattached line ends would be exactly sqrt 3"
long. Additionally, the smallest angle would be exactly 30 degrees, the
larger angle would be exactly 60 degrees and the biggest angle would be
exactly 90 degrees.

--
PDQ

--

"Enoch Root" wrote in message ...
| PDQ wrote:
| "Guess who" wrote in message ...
| | On Mon, 10 Apr 2006 15:07:38 GMT, "Gooey TARBALLS"
| | wrote:
| |
| | Draw a vertical line desired length of shortest leg (x).
| | Draw a horizontal line 2x in length from the same point.
| |
| | Could the result be that 30,60, 90 degree triangle? (if one connected the
| | points?
| |
| | NO. As was pointed out the two arms [at right angles] of a 30,60
| | right triangle are in the ratio 1:sqrt(3). The Hypotenuse is the side
| | twice the shorter.
| |
|
| You sure about this??
|
| Seems to me the ratio of the sides of a scalene right triangle (30,60,90) is
|
| 1:2:sqrt3.
|
| and, according to Pythagorus,
|
| The square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other 2 sides.
|
| Give all this balderdash, were one to lay out a horizontal base line of 2" and erect a vertical line at one end of the base line that is 1" in height, the distance between the two unattached line ends would be exactly sqrt 3" long. Additionally, the smallest angle would be exactly 30 degrees, the larger angle would be exactly 60 degrees and the biggest angle would be exactly 90 degrees.
|
| So, you're saying 1^2 + 2^2 = 3 ?
|
| er
| --
| email not valid

There are days when there seems to be a dis-connection between my extremities.

Must be age related --- I meant sqrt 5 and my fingers weren't listening.

To get a side of sqrt 3 one would need the hypotenuse to be exactly 2" long.

The height of an iscoceles triangle of this side length would be sqrt 3.
--
PDQ

--

On Mon, 10 Apr 2006 23:58:00 GMT, "CW" wrote:

Nope.

"PDQ" wrote in message
.. .
Give all this balderdash, were one to lay out a horizontal base line of 2"
and erect a vertical line at one end of the base line that is 1" in height,
the distance between the two unattached line ends would be exactly sqrt 3"
long.

NO. The line joining those ends is the hypotenuse. Never mind the
other relationships, but consider that the hypotenuse is *always* the
longest side of a right triangle [very simple to prove]. Drag out
your calculator if you have to, but sqrt(3) is about 1.7321, which is
still less than 2 where I come from.

In your diagram the length would be sqrt(1^2 + 2^2) = sqrt(5), and the
angles would not be 30,60.

Andy Dingley wrote:
On Mon, 10 Apr 2006 03:16:56 GMT, "Gooey TARBALLS"
wrote:

Perhaps you would care to post your solution here?

In ASCII ? Not really !

My searching seemed to indicate it was a Trig problem.

You can do it either way. For a general solution to any angle, then trig
would be the way to go - especially with cheap calculators. However
90°. 60° and 30° are easy constructions with just dividers and they're
worth knowing.

Draw a circle. With your dividers set on the same radius, step out this
distance repeatedly around the circumference of the circle. It fits
exactly 6 times, meaning that the radii to each of these points make
angles of 60°.

To get from 60° to 30°, then you can either bisect it or you can
subtract 60° from 90° (you'll often already have a right angle laid out
for some other purpose).

To bisect the angle, use the dividers as before and mark out a point at
approximately two radii from the centre, at the intersection of arcs
drawn from two of the circumferential points you marked earlier. A
radius drawn out to this point bisects the circumferential points, at an
angle of 30°


I must be lazy. After getting 2 points on the circle, which, combined
with the center point, gives me a 60° angle and a side of 1 unit, I
extended the other side of the angle to 2 units. Connecting this
endpoint with the end of the 1 unit side, on the circle, gives me the
side that's the square root of 3 units.
Joe

In article ,
Robert Bonomi wrote:
In article ,
Australopithecus scobis wrote:
On Mon, 10 Apr 2006 03:16:56 +0000, Gooey TARBALLS opined:

Perhaps you would care to post your solution here?

Draw a long line (1). Construct a perpendicular (2). Set dividers to
[[.. munch rest of a detailed description ..]]

*SIGH* There's an even easier way.

[[.. sneck ..]]

and I was still doing it the hard way.

the _easy_ way:

1) Draw a straight line.
2) grab a compass, set it to any convenient value.
3) put the point on the line, and draw a 180-degree arc, where both ends
are on the line.
4) put the compass point on one end of the arc from step 3
5) strike a new arc, intersecting the arc from step 3
6) connect that intersection point to the two ends of the 180-degree arc.

Voila! a 30-60-90 right triangle.

On Mon, 10 Apr 2006 23:08:30 +0000, Robert Bonomi opined:

ABF is a 30-60-90 right triangle.

It takes _way_ longer to describe it than to do it. grin

The point, of course, is that 30:60:90 triangles are trivial to generate
with geometry. OP has his pick, so far, of at least three different
methods.

I just use a plastic drafting triangle.
--
"Keep your ass behind you"
wreck20051219 at spambob.net